Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int rd() {
  int res = 0, c;
  for (c = getchar(); !isdigit(c); c = getchar())
    ;
  for (; isdigit(c); c = getchar()) res = (res << 1) + (res << 3) + (c ^ 48);
  return res;
}
struct edge {
  int to, nxt;
} e[100005 << 1];
int n, m, q, x, y, a, head[100005];
void link(int s, int t, int num) {
  e[num] = (edge)<% t, head[s] %>;
  head[s] = num;
}
vector<int> c[100005];
int top[100005], son[100005], fa[100005], sz[100005], id[100005], od[100005],
    dep[100005], tot;
void dfs1(int x, int f) {
  sz[x] = 1, dep[x] = dep[f] + 1, fa[x] = f;
  for (register int i = head[x]; i; i = e[i].nxt) {
    int y = e[i].to;
    if (y == f) continue;
    dfs1(y, x);
    sz[x] += sz[y];
    if (sz[son[x]] < sz[y]) son[x] = y;
  }
}
void dfs2(int x, int tp) {
  id[x] = ++tot, od[tot] = x;
  top[x] = tp;
  if (son[x]) dfs2(son[x], tp);
  for (register int i = head[x]; i; i = e[i].nxt) {
    int y = e[i].to;
    if (y == fa[x] || y == son[x]) continue;
    dfs2(y, y);
  }
}
void Upd(int *L, int *R, int l1, int l2, int *a, int &sz) {
  int i = 0, j = 0;
  sz = 0;
  while (i < l1 && j < l2 && sz < 10)
    if (L[i] < R[j])
      a[sz++] = L[i], i++;
    else
      a[sz++] = R[j], j++;
  while (i < l1 && sz < 10) a[sz++] = L[i], i++;
  while (j < l2 && sz < 10) a[sz++] = R[j], j++;
}
struct node {
  int l, r, sz;
  int a[10];
} t[100005 << 2];
void build(int p, int l, int r) {
  t[p].l = l, t[p].r = r;
  if (l == r) {
    int *a = t[p].a;
    sort(c[od[l]].begin(), c[od[l]].end());
    t[p].sz = c[od[l]].size();
    t[p].sz = min(t[p].sz, 10);
    for (register int i = 0; i < t[p].sz; i++) a[i] = c[od[l]][i];
    return;
  }
  int mid = l + r >> 1;
  build(p << 1, l, mid);
  build(p << 1 | 1, mid + 1, r);
  int &sz = t[p].sz, *a = t[p].a;
  int l1 = t[p << 1].sz, l2 = t[p << 1 | 1].sz;
  int *L = t[p << 1].a, *R = t[p << 1 | 1].a;
  Upd(L, R, l1, l2, a, sz);
}
int q_cnt(int p, int l, int r) {
  if (t[p].l == l && t[p].r == r) return t[p].sz;
  int mid = t[p].l + t[p].r >> 1;
  if (r <= mid) return q_cnt(p << 1, l, r);
  if (l > mid) return q_cnt(p << 1 | 1, l, r);
  return q_cnt(p << 1, l, mid) + q_cnt(p << 1 | 1, mid + 1, r);
}
int w[15], cnt;
void query(int p, int l, int r) {
  if (t[p].l == l && t[p].r == r) {
    int tmp[15], len = 0;
    Upd(w, t[p].a, cnt, t[p].sz, tmp, len);
    for (int i = 0; i < len; i++) w[i] = tmp[i];
    cnt = len;
    return;
  }
  int mid = t[p].l + t[p].r >> 1;
  if (r <= mid)
    query(p << 1, l, r);
  else if (l > mid)
    query(p << 1 | 1, l, r);
  else {
    query(p << 1, l, mid);
    query(p << 1 | 1, mid + 1, r);
  }
}
void calc(int x, int y, int a) {
  cnt = 0;
  while (top[x] != top[y]) {
    if (dep[top[x]] > dep[top[y]]) {
      query(1, id[top[x]], id[x]);
      x = fa[top[x]];
    } else {
      query(1, id[top[y]], id[y]);
      y = fa[top[y]];
    }
  }
  if (dep[x] < dep[y])
    query(1, id[x], id[y]);
  else
    query(1, id[y], id[x]);
  a = min(a, cnt);
  printf("%d", a);
  for (int i = 0; i < a; i++) printf(" %d", w[i]);
}
int main() {
  n = rd(), m = rd(), q = rd();
  for (register int i = 1; i < n; i++) {
    x = rd(), y = rd();
    link(x, y, (i << 1) - 1);
    link(y, x, i << 1);
  }
  for (register int i = 1; i <= m; i++) c[rd()].push_back(i);
  dfs1(1, 0);
  dfs2(1, 1);
  build(1, 1, n);
  for (register int i = 1; i <= q; i++) {
    x = rd(), y = rd(), a = rd();
    calc(x, y, a);
    if (i != q) puts("");
  }
}